2 Answers
2

There isn't a unique answer, especially if (infinite) series are used. But if, as the example formula indicates, you want a polynomial in 1/z, then the result is unique up to multiplying the numerator and denominator by a scalar factor.

Since Numerator@f == (Denominator@f /. {h2 -> h1, w2 -> w1}) it should be suficient to have only the series decomposition of the numerator, no? Or are you aiming for something else?
Series[Numerator@f, {z, 0, 2}]

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